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The corresponding sides of $\Delta ABC$ and $\Delta PQR$ are in the ratio $3 : 5$. $AD \perp BC$ and $PS \perp QR$ as shown in the following figures : (i) Prove that $\Delta ADC \sim \Delta PSR$ (ii) If $AD = 4$ cm, find the length of $PS$. (iii) Using (ii) find $ar (\Delta ABC) : ar (\Delta PQR)$
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As, $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR} = \frac{3}{5}$
$\implies \Delta ABC \sim \Delta PQR \implies \angle C = \angle R$
(i) In $\Delta ADC$ and $\Delta PSR, \angle ADC = \angle PSR = 90^{\circ}$ and $\angle C = \angle R \implies \Delta ADC \sim \Delta PSR$
(ii) $\frac{AD}{PS} = \frac{AC}{PR} = \frac{3}{5} \implies \frac{4}{PS} = \frac{3}{5} \implies PS = \frac{20}{3}$ cm
(iii) $\frac{ar (\Delta ABC)}{ar (\Delta PQR)} = \frac{\frac{1}{2} \times BC \times AD}{\frac{1}{2} \times QR \times PS} = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25} \implies ar (\Delta ABC) : ar (\Delta PQR) = 9 : 25$
$\implies \Delta ABC \sim \Delta PQR \implies \angle C = \angle R$
(i) In $\Delta ADC$ and $\Delta PSR, \angle ADC = \angle PSR = 90^{\circ}$ and $\angle C = \angle R \implies \Delta ADC \sim \Delta PSR$
(ii) $\frac{AD}{PS} = \frac{AC}{PR} = \frac{3}{5} \implies \frac{4}{PS} = \frac{3}{5} \implies PS = \frac{20}{3}$ cm
(iii) $\frac{ar (\Delta ABC)}{ar (\Delta PQR)} = \frac{\frac{1}{2} \times BC \times AD}{\frac{1}{2} \times QR \times PS} = \frac{3}{5} \times \frac{3}{5} = \frac{9}{25} \implies ar (\Delta ABC) : ar (\Delta PQR) = 9 : 25$